So... hmm bear in mind, when the boat goes upstream, it goes against the stream, so, if the boat has speed rate of say "b", and the stream has a rate of "r", then the speed going up is b - r, the boat's rate minus the streams, because the stream is subtracting speed as it goes up
going downstream is a bit different, the stream speed is "added" to boat's
so the boat is really going faster, is going b + r
notice, the distance is the same, upstream as well as downstream
thus
![\bf \begin{cases} b=\textit{rate of the boat}\\ r=\textit{rate of the river} \end{cases}\qquad thus \\\\\\ \begin{array}{lccclll} &distance&rate&time(hrs)\\ &----&----&----\\ upstream&48&b-r&4\\ downstream&48&b+4&3 \end{array} \\\\\\ \begin{cases} 48=(b-r)(4)\to 48=4b-4r\\\\ \frac{48-4b}{-4}=r\\ --------------\\ 48=(b+r)(3)\\ -----------------------------\\\\ thus\\\\ 48=\left[ b+\left(\boxed{\frac{48-4b}{-4}}\right) \right] (3) \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0Ab%3D%5Ctextit%7Brate%20of%20the%20boat%7D%5C%5C%0Ar%3D%5Ctextit%7Brate%20of%20the%20river%7D%0A%5Cend%7Bcases%7D%5Cqquad%20thus%0A%5C%5C%5C%5C%5C%5C%0A%0A%5Cbegin%7Barray%7D%7Blccclll%7D%0A%26distance%26rate%26time%28hrs%29%5C%5C%0A%26----%26----%26----%5C%5C%0Aupstream%2648%26b-r%264%5C%5C%0Adownstream%2648%26b%2B4%263%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0A%0A%5Cbegin%7Bcases%7D%0A48%3D%28b-r%29%284%29%5Cto%2048%3D4b-4r%5C%5C%5C%5C%0A%5Cfrac%7B48-4b%7D%7B-4%7D%3Dr%5C%5C%0A--------------%5C%5C%0A48%3D%28b%2Br%29%283%29%5C%5C%0A-----------------------------%5C%5C%5C%5C%0Athus%5C%5C%5C%5C%0A48%3D%5Cleft%5B%20b%2B%5Cleft%28%5Cboxed%7B%5Cfrac%7B48-4b%7D%7B-4%7D%7D%5Cright%29%20%5Cright%5D%20%283%29%0A%5Cend%7Bcases%7D)
solve for "r", to see what the stream's rate is
what about the boat's? well, just plug the value for "r" on either equation and solve for "b"
Answer:
−7(2x−11)
use distributive property
-7(2x)-7(-11)
-14x+77
Answer:
lila added 1/12 more than fiona
Step-by-step explanation:
find GCF (12), multiply amounts added accordingly
lila added 9/12 (multiply 3/4 by 3)
fiona added 2/3 (multiply 2/3 by 4)
Answer:

Step-by-step explanation:
Given function:

where:
- h = height (in feet)
- t = times (in seconds)
Rewrite in the form 
The quickest way to do this is to expand
and compare coefficients with the original function.

Comparing coefficients with 



Therefore, substituting the found values into the equation:

Sorry I'm gonna take a guess.......... d?